The first thing i did was i assumed the pair of numbers as 75a,75b where a,b are co-prime.
Now a,b have to be factors of 240 since 18000/75 = 240 .The prime factorization of 240=2^4*3*5
I consider one of the factor to be 1 and the other factor will be all the factors of 240(excluding 1) which is 5*2*2-1=19
i) one of the factor is factors of 2^4 (excluding 1) and the other is factors of 3*5(excluding 1) which is 4 ways multiplied by 3 ways = 12
ii) similarly one factor is factors of 3 (excluding 1) and the other factor is factors of 2^4*5 (excluding 1) which is 1 way multiplied by 9 ways = 9
iii) similarly we get 9 ways again when one of the factor is 5.
Now i think in case 2 all the cases are counted twice so the total number of cases is actually (12+9+9)/2=15
so total pairs of distinct factors whose HCF is 75 is equal to 15+19=34.
Now i dont know if the answer is correct or not. Also any suggestions for a more efficient method is appreciated.
